# The function ‘t’ which maps temperature in degree Celsius into temperature in degree

Question:

The function ' $t$ ' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(\mathrm{C})=\frac{9 \mathrm{C}}{5}+32$.

Find

(i) $t$ ( 0 )

(ii) $t$ (28)

(iii) $t(-10)$

(iv) The value of $\mathrm{C}$, when $t(\mathrm{C})=212$

Solution:

The given function is $t(\mathrm{C})=\frac{9 \mathrm{C}}{5}+32$.

Therefore,

(i) $t(0)=\frac{9 \times 0}{5}+32=0+32=32$

(ii) $t(28)=\frac{9 \times 28}{5}+32=\frac{252+160}{5}=\frac{412}{5}$

(iii) $t(-10)=\frac{9 \times(-10)}{5}+32=9 \times(-2)+32=-18+32=14$

(iv) It is given that $t(\mathrm{C})=212$

$\therefore 212=\frac{9 \mathrm{C}}{5}+32$

$\therefore 212=\frac{9 C}{5}+32$

$\Rightarrow \frac{9 C}{5}=180$

$\Rightarrow 9 C=180 \times 5$

$\Rightarrow C=\frac{180 \times 5}{9}=100$

Thus, the value of $t$, when $t(C)=212$, is 100 .

$\Rightarrow \frac{9 C}{5}=212-32$